Answer
a) $W=-4.49\times10^{11}J$
b) $F=2.49\times10^5N$
Work Step by Step
a) From the given information, $v_0=7100m/s$, $v_f=5500m/s$ and $m_{asteroid}=4.5\times10^4kg$. Using these, we can calculate the kinetic energies:
$KE_0=\frac{1}{2}mv_0^2=1.13\times10^{12}J$
$KE_f=\frac{1}{2}mv_f^2=6.81\times10^{11}J$
According to the work-energy theorem, the work done by the force is $W=KE_f-KE_0=-4.49\times10^{11}J$
b) We have $$W=(F\cos\theta)s$$
We know $s=1.8\times10^6m$, and because the force acts along the asteroid's displacement but slows it down, it goes in the opposite direction, so $\theta=180^o$
$$F=\frac{W}{\cos180 s}=\frac{W}{-s}=2.49\times10^5N$$
